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Question

(D) Let the two light sources be $A$ and $B$ with equal luminous intensity $I$. Let the screen be placed at a distance $x$ from source $A$. Then the d

Vedclass · Accepted Answer

Both $(b)$ and $(c)$

Vedclass · Answer

(D) Let the two light sources be $A$ and $B$ with equal luminous intensity $I$. Let the screen be placed at a distance $x$ from source $A$. Then the distance from source $B$ is $(1.2 - x)$.The illuminance $E$ at a distance $r$ from a source of intensity $I$ is given by $E = I/r^2$.According to the problem,the illuminance on one face is four times the illuminance on the other face:$\frac{I}{x^2} = 4 	imes \frac{I}{(1.2 - x)^2}$ or $\frac{I}{(1.2 - x)^2} = 4 	imes \frac{I}{x^2}$.Case $1$: $\frac{1}{x^2} = \frac{4}{(1.2 - x)^2}$Taking the square root on both sides: $\frac{1}{x} = \frac{2}{1.2 - x}$$1.2 - x = 2x \Rightarrow 3x = 1.2 \Rightarrow x = 0.4\, m$.Case $2$: $\frac{1}{(1.2 - x)^2} = \frac{4}{x^2}$Taking the square root on both sides: $\frac{1}{1.2 - x} = \frac{2}{x}$$x = 2(1.2 - x) \Rightarrow x = 2.4 - 2x \Rightarrow 3x = 2.4 \Rightarrow x = 0.8\, m$.Thus,the screen can be placed at $0.4\, m$ or $0.8\, m$ from source $A$. Therefore,the correct option is $(d)$.

Two light sources with equal luminous intensity are placed at a distance of $1.2\, m$ from each other. Where should a screen be placed between them such that the illuminance on one of its faces is four times that on the other face?

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